# For a frequency distribution, the mean is 200, the coefficient of variation is 8% and Karl Pearsonian’s coefficient of skewness is 0.3. Find the mode and median of the distribution. - Mathematics and Statistics

Sum

For a frequency distribution, the mean is 200, the coefficient of variation is 8% and Karl Pearsonian’s coefficient of skewness is 0.3. Find the mode and median of the distribution.

#### Solution

Mean = bar"x" = 200, Coefficient of variation,
C.V. = 8%, Skp = 0.3

C.V. = sigma/bar"x" xx 100, where σ = standard deviation

∴ 8 = sigma/200 xx 100

∴ sigma = (8 xx 200)/100 = 16

Now, Skp = ("Mean"-"Mode")/"S.D."

∴ 0.3 = (200 - "Mode")/16

∴ 0.3 × 16 = 200 – Mode
∴ Mode = 200 – 4.8 = 195.2
Since, Mean – Mode = 3(mean – Median)
∴ 200 – 195.2 = 3(200 – Median)
∴ 4.8 = 600 – 3 Median
∴ 3 Median = 600 – 4.8 = 595.2

∴ Median = (595.2)/3 = 198.4

Concept: Measures of Skewness - Karl Pearson’S Coefficient of Skewness (Pearsonian Coefficient of Skewness)
Is there an error in this question or solution?
Chapter 3: Skewness - Miscellaneous Exercise 3 [Page 44]

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